Question

The following thermodynamic data are available for octane, oxygen gas, carbon dioxide gas, water, and water vapor: Molecule ΔH∘f (kJ/mol) C8H18(l) −250.1 O2(g) 0 CO2(g) −393.5 H2O(l) −285.8 H2O(g) −241.8 Part B Calculate ΔHrxn for the combustion of octane by using enthalpies of formation from the transition above. Express the energy in kilojoules per mole to three significant figures. ΔHrxn Δ H r x n = nothing kJ/mol

Answer 1

Answer : The $\Delta H_{rxn}$ for the combustion of octane in kilojoules per mole is, $5.35\times 10^3kJ/mol$

Explanation :

Enthalpy change : It is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as $\Delta H^o$

The equation used to calculate enthalpy change is of a reaction is:  

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)]$

The chemical reaction for the combustion of octane follows the equation:

$2C_{8}H_{18}(l)+25O_2(g)\rightarrow 16CO_2(g)+18H_2O(l)$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(16\times \Delta H^o_f_{(CO_2(g))})+(18\times \Delta H^o_f_{(H_2O(l))})]-[(2\times \Delta H^o_f_{(C_8H_{18}(l))})+(25\times \Delta H^o_f_{(O_2(g))})]$

We are given:

$\Delta H^o_f_{(C_8H_{18}(l))}=-250.1kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_f_{(H_2O(l))}=-285.8kJ/mol$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(16\times (-393.5))+(18\times (-285.8))]-[(2\times (-250.1))+(25\times (0))]=-1.07\times 10^4kJ$

Now we have to calculate the $\Delta H_{rxn}$ for the combustion of octane in kilojoules per mole.

From the reaction we conclude that,

As, 2 moles of compound released heat = $1.07\times 10^4kJ$

So, 1 moles of compound released heat = $\frac{1.07\times 10^4kJ}{2}=5.35\times 10^3kJ/mol$

Therefore, the $\Delta H_{rxn}$ for the combustion of octane in kilojoules per mole is, $5.35\times 10^3kJ/mol$

Answer 2

Answer: The enthalpy change of the reaction comes out to be $1.01\times 10^4kJ/mol$

Explanation:

The chemical reaction for the combustion of octane follows the equation:

$2C_{8}H_{18}(l)+25O_2(g)\rightarrow 16CO_2(g)+18H_2O(g)$

The equation used to calculate enthalpy change is of a reaction is:  

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)]$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(16\times \Delta H^o_f_{(CO_2(g))})+(18\times \Delta H^o_f_{(H_2O(g))})]-[(2\times \Delta H^o_f_{(C_8H_{18}(l))})+(25\times \Delta H^o_f_{(O_2(g))})]$

We are given:

$\Delta H^o_f_{(C_8H_{18}(l))}=-250.1kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_f_{(H_2O(g))}=-241.8kJ/mol$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(16\times (-393.5))+(18\times (-241.8))]-[(2\times (-250.1))+(25\times (0))]=10148.2kJ/mol=1.01\times 10^4kJ/mol$

Hence, the enthalpy change of the reaction comes out to be $1.01\times 10^4kJ/mol$